Dada una array arr[] de tamaño N y un número entero K , la tarea es encontrar el número de subsecuencias de longitud K de esta array tal que la suma de estas subsecuencias sea la mínima posible.
Ejemplos:
Entrada: arr[] = {1, 2, 3, 4}, K = 2
Salida: 1
Las subsecuencias de longitud 2 son (1, 2), (1, 3), (1, 4),
(2, 3) , (2, 4) y (3, 4).
La suma mínima es 3 y la única subsecuencia
con esta suma es (1, 2).Entrada: arr[] = {2, 1, 2, 2, 2, 1}, K = 3
Salida: 4
Enfoque: La suma mínima posible de una subsecuencia de longitud K de la array dada es la suma de los K elementos más pequeños de la array. Sea X el elemento máximo entre los K elementos más pequeños del arreglo, y sea el número de veces que ocurre entre los K, los elementos más pequeños del arreglo, Y , y, su ocurrencia total, en el arreglo completo, sea cntX . Ahora, hay cntX C Y formas de seleccionar este elemento, en los K elementos más pequeños, que es el conteo de subsecuencias requeridas.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the value of
// Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
int C[n + 1][k + 1];
int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= min(i, k); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
return C[n][k];
}
// Function to return the count
// of valid subsequences
int cntSubSeq(int arr[], int n, int k)
{
// Sort the array
sort(arr, arr + n);
// Maximum among the minimum K elements
int num = arr[k - 1];
// Y will store the frequency of num
// in the minimum K elements
int Y = 0;
for (int i = k - 1; i >= 0; i--) {
if (arr[i] == num)
Y++;
}
// cntX will store the frequency of
// num in the complete array
int cntX = Y;
for (int i = k; i < n; i++) {
if (arr[i] == num)
cntX++;
}
return binomialCoeff(cntX, Y);
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int n = sizeof(arr) / sizeof(int);
int k = 2;
cout << cntSubSeq(arr, n, k);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the value of
// Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
int C[][] = new int [n + 1][k + 1];
int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.min(i, k); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] +
C[i - 1][j];
}
}
return C[n][k];
}
// Function to return the count
// of valid subsequences
static int cntSubSeq(int arr[], int n, int k)
{
// Sort the array
Arrays.sort(arr);
// Maximum among the minimum K elements
int num = arr[k - 1];
// Y will store the frequency of num
// in the minimum K elements
int Y = 0;
for (int i = k - 1; i >= 0; i--)
{
if (arr[i] == num)
Y++;
}
// cntX will store the frequency of
// num in the complete array
int cntX = Y;
for (int i = k; i < n; i++)
{
if (arr[i] == num)
cntX++;
}
return binomialCoeff(cntX, Y);
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1, 2, 3, 4 };
int n = arr.length;
int k = 2;
System.out.println(cntSubSeq(arr, n, k));
}
}
// This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the value of
// Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k)
{
int [,]C = new int [n + 1, k + 1];
int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= Math.Min(i, k); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i, j] = 1;
// Calculate value using previously
// stored values
else
C[i, j] = C[i - 1, j - 1] +
C[i - 1, j];
}
}
return C[n, k];
}
// Function to return the count
// of valid subsequences
static int cntSubSeq(int []arr, int n, int k)
{
// Sort the array
Array.Sort(arr);
// Maximum among the minimum K elements
int num = arr[k - 1];
// Y will store the frequency of num
// in the minimum K elements
int Y = 0;
for (int i = k - 1; i >= 0; i--)
{
if (arr[i] == num)
Y++;
}
// cntX will store the frequency of
// num in the complete array
int cntX = Y;
for (int i = k; i < n; i++)
{
if (arr[i] == num)
cntX++;
}
return binomialCoeff(cntX, Y);
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 1, 2, 3, 4 };
int n = arr.Length;
int k = 2;
Console.WriteLine(cntSubSeq(arr, n, k));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach # Function to return the value of # Binomial Coefficient C(n, k) def binomialCoeff(n, k) : C = [[0 for i in range(n + 1)] for j in range(k + 1)] # Calculate value of Binomial Coefficient # in bottom up manner for i in range (0, n + 1 ): for j in range (0, min(i, k) + 1): # Base Cases if (j == 0 or j == i): C[i][j] = 1 # Calculate value using previously # stored values else : C[i][j] = C[i - 1][j - 1] + C[i - 1][j] return C[n][k] # Function to return the count # of valid subsequences def cntSubSeq(arr, n, k) : # Sort the array arr.sort() # Maximum among the minimum K elements num = arr[k - 1]; # Y will store the frequency of num # in the minimum K elements Y = 0; for i in range (k - 1, -1, 1) : if (arr[i] == num): Y += 1 # cntX will store the frequency of # num in the complete array cntX = Y; for i in range (k, n): if (arr[i] == num) : cntX += 1 return binomialCoeff(cntX, Y) # Driver code arr = [ 1, 2, 3, 4 ] n = len(arr) k = 2 print(cntSubSeq(arr, n, k)) # This code is contributed by ihritik
Javascript
<script>
// Javascript implementation of the
// above approach
// Function for the binomial coefficient
function binomialCoeff(n, k)
{
var C = new Array(n + 1);
// Loop to create 2D array using 1D array
for (var i = 0; i < C.length; i++) {
C[i] = new Array(k + 1);
}
var i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= Math.min(i, k); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
return C[n][k];
}
// Function to return the count
// of valid subsequences
function cntSubSeq(arr, n, k)
{
// Sort the array
arr.sort();
// Maximum among the minimum K elements
var num = arr[k - 1];
// Y will store the frequency of num
// in the minimum K elements
var Y = 0;
for (var i = k - 1; i >= 0; i--) {
if (arr[i] == num)
Y+=1;
}
// cntX will store the frequency of
// num in the complete array
var cntX = Y;
for (var i = k; i < n; i++) {
if (arr[i] == num)
cntX+=1;
}
return binomialCoeff(cntX, Y);
}
// Driver code
var arr = [ 1, 2, 3, 4 ];
var n = arr.length;
var k = 2;
document.write(cntSubSeq(arr, n, k));
// This code is contributed by ShubhamSingh10
</script>
Producción:
1
Complejidad temporal: o(n 2 )
Espacio Auxiliar: O(n * k)